Reference: Wells- Differential analysis on complex manifolds p.69
Consider a complex vector bundle $E\rightarrow X$.
Let $\mathscr{E}(E)$ denote the collection of smooth sections and let $\mathscr{E}^k$ denote the smooth $k$-forms on $X$ and let $\mathscr{E}^k(E)$ denote smooth $k$-forms on $E$. Then, we have an isomorphism $\mathscr{E}^k\otimes_{C^\infty} \mathscr{E}(E)\cong \mathscr{E}^p(E)$.
Let's denote the image of $\phi\otimes \xi$ under the isomorphism by $\phi\cdot \xi$ for $\phi\in \mathscr{E}^k$ and $\xi\in \mathscr{E}(E)$.
Let $U$ be open in $X$ and $f=\{e_1,...,e_r\}$ be a frame for $E$ over $U$.
Then, we have a local representation for an $E$-valued $k$-form $\xi$ (locally defined on $U$) given by $(\xi^1(f),...,\xi^r(f))$, defined by the relation $\xi=\sum_{\rho=1}^r \xi^\rho(f)\cdot e_\rho$.
Namely, let $x\in U$ and let $(\omega_1,....,\omega_s)$ be a frame for $\wedge^k T^*(X)\otimes \mathbb{C}$ at $x$.
Then, we can write $\xi(x)=\sum_{\rho,j} \phi_{j\rho}(x)\omega_j(x)\otimes e_\rho(x)$ where the $\phi_{j\rho}$ are uniquely determined $C^\infty$ functions defined near $x$.
I'm fine with what illustrated above, but I don't get the next line:
"Let $\xi^k=\sum_j \phi_{j\rho}\omega_j$, and it is easy to check that the differential form $\xi^\rho$ so determined is independent of the choice of frame $(\omega_1,...,\omega_s)$."
I think it should be $\xi^\rho$ at first instead of $\xi^k$, am I correct? And of course from the definition of $\xi^\rho$, it only depends on the choice of $\{e_1,...,e_r\}$. So there is nothing to check here, but why the author wrote "it is easy to check..."?
Am I missing something?
Thank you in advance!